Free AI LeetCode Communication Coach — Binary Search

The problem

Given a sorted ascending array and a target, return the index or -1. Aim for O(log n).

Example: nums = [1,3,5,7,9], target = 7 → 3

Say the discard line every time.

left, right = 0, len(nums) - 1
while left <= right:
    mid = (left + right) // 2
    if nums[mid] == target:
        return mid
    if nums[mid] < target:
        left = mid + 1
    else:
        right = mid - 1
return -1

How to explain it

1. 1. Restate

   Say the problem in your own words.

   One or two sentences. Show you understood the input, the output, and the goal — not that you memorised the prompt.

2. 2. Approach

   Name the method before you code.

   Brute force first if you need it, then the structure you will use: hash map, two pointers, stack, binary search.

3. 3. Example

   Walk one concrete input.

   Pick small numbers. Say what you store, what you compare, and what you return. Interviewers follow an example more easily than abstract talk.

4. 4. Time and space

   One sentence each.

   After the example, before you claim you are done. “Time is O(n) because we scan once. Space is O(n) for the map.”

5. 5. Edge cases

   Name at least one unusual input.

   Empty input, duplicates, already sorted, overflow. Invite a follow-up: “I would also check …”

• “Because the array is sorted I can discard half each step.”
• “If the mid is too small, I move left up.”

Practise out loud

1. 1. Restate

   The array is sorted. I need the index of the target, or minus one.

2. 2. Approach

   Because it is sorted I can binary search. I keep left and right, look at mid, and discard the half that cannot contain the target.

3. 3. Example

   On 1,3,5,7,9 target 7 — mid is 5, too small, so I search the right half and then hit 7.

4. 4. Time and space

   Time is O(log n). Space is O(1).

5. 5. Edge cases

   I would check an empty array and a target smaller than the first element.

Common mistakes when explaining Binary Search

• Skipping the “I discard this half” sentence.
• Off-by-one on left/right without narrating the loop invariant.